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## Physics library

### Course: Physics library > Unit 1

Lesson 5: Old videos on projectile motion# Projectile motion (part 3)

An example of solving for the final velocity when you know the change in distance, time, initial velocity, and acceleration. Created by Sal Khan.

## Want to join the conversation?

- What is the difference between constant acceleration and constant velocity?(5 votes)
- constant acceleration means the velocity is changing at a constant rate it can increase or decrease but constant velocity means that if a particle is having constant velocity it will move with that velocity only until and unless any external force is applied to it(9 votes)

- Are we supposed to memorize each and every formula for time? I need to know how to get time when given different values ( distance, initial velocity, final velocity, displacement, ect.) Can you give me a list of formulas to use in each scenario? Thanks!!!(4 votes)
- it's quite simple to remember.

1. s=ut+1/2at^2

2. v^2=u^2+2as

3. v=u+at

4.t=v-u/a

where,v is final velocity,u initial velocity,t the time,a acceleration and s the distance.(4 votes)

- Hi Sal,

You have taken upwards positive and downwards negative. Could we take upwards negative and downwards positive?(2 votes)- Yes, since coordinates have no effect on physics.(4 votes)

- At1:45Salman Khan told that Avg. Velocity = Change in distance by change in time, But I thought that it was Initial vleocity+ Final Velocity /2? Which one is correct?(3 votes)
- Your formula only works when acceleration is constant.(3 votes)

- Isn't sal supposed to use displacement (delta S) instead of distance(1 vote)
- In7:40he stated that the formula for average velocity is (Vf+Vi)/2 assuming the acceleration is constant , but what if the acceleration wasnt constant and we were dealing with a jerky motion what the formula for average velocity might be ?(1 vote)
- Then you would need to calculate the average in a different way, perhaps with calculus.(1 vote)

- My problem has angles and no velocity is given. How do I figure it out when I'm given an angle and how far down the projectile went from the horizon?(1 vote)
- i am unable to do the problems in our text using the formula Sal says?what should i do?(1 vote)
- When we say "change in displacement 'delta S' ", do we mean the change in the displacement form a particular reference point?(3 votes)
- displacement itself is change in position vector. final minus initial. initial can be your reference. i hope you should not say change in displacement.(0 votes)

- Aren't we looking for the time? So why is the last formula is for distance? I tried my best to derive the time from D = ViT + ( ( aT^2 ) / 2 ) but no luck. I was hoping Sal would give the formula for time with the given distance, initial velocity and acceleration. Seems impossible since we need time to determine final velocity.(0 votes)
- transpose everything but time to LHS to find time(2 votes)

## Video transcript

In the last video, I said that
we started off with the change in distance, so we
said that we know the change in distance. These are the things
that we are given. We're given the acceleration,
we're given the initial velocity, and I asked you how
do we figure out what the final velocity is? In the last video-- if you don't
remember it, go watch that last video again-- we
derived the formula that vf squared, the final velocity
squared, is equal to the initial velocity squared
plus 2 times the change in distance. You'll sometimes just see it
written as 2 times distance, because we assume that the
initial distance is at point 0, so the change in distance
would just be the final distance. We could write it either way,
and hopefully, at this point, you see why I keep switching
between change in distance and distance. It's just so you're comfortable
when you see it either way. This is for the situation
when we didn't know what the vf was. Let's say we want to solve
for time instead. Once we solve for the final
velocity, we could actually solve for time, and I'll show
you how to do that, but let's say we didn't want to go through
this step-- how can we solve for time directly, given
the change in distance, the acceleration, and the
initial velocity? Let's go back once again to
the most basic distance formula-- not the distance
formula, but how distance relates to velocity. We know that-- I'll write it
slightly different this time-- the change in distance over the
change in time is equal to the average velocity. We could have rewritten this as
the change in distance is equal to the average velocity
times the change in time. This is change in time and
change in distance. Sometimes we'll just see this
written as d equals-- let me write this in a different
color, so we have some variety-- velocity times time,
or d equals rate times time. The reason why I have change in
distance here, or change in time, is that I'm not assuming
necessarily that we're starting off at the point
0 or at time 0. If we do, then it just turns out
to the final distance is equal to the average velocity
times the final time, but let's stick to this. We want to figure out time
given this set of inputs. Let's go from this equation. If we want to solve for time,
or the change in time, we could just could divide both
sides by the average velocity-- actually, no,
let's not do that. Let's just stay in terms
of change in distance. I've wasted space too fast,
so let me clear this and start again. We're given change in distance,
initial velocity, and acceleration, and we want
to figure out what the time is-- it's really the change in
time, but let's just assume that we start time 0, so it's
kind of the final time. Let's just start with the simple
formula: distance, or change in distance-- I'll use
them interchangeably, with a lower case d this time-- is
equal to the average velocity times time. What's the average velocity? The average velocity is just the
initial velocity plus the final velocity over 2. The only reason why we can just
average the initial and the final is because we're
assuming constant acceleration, and that's very
important, but in most projectile problems, we do have
constant acceleration-- downwards-- and that's
gravity. We can assume, and we can do
this-- we can say that the average of the initial and the
final velocity is the average velocity, and then we multiply
that times time. Can we use this equation
directly? No. we know acceleration, but
don't know final velocity. If we can write this final
velocity in terms of the other things in this equation, then
maybe we can solve for time. Let's try to do that: distance
is equal to-- let me take a little side here. What do we know about
final velocity? We know that the change in
velocity is equal to acceleration times time,
assuming that time starts a t equals 0. The change in velocity is the
same thing is vf minus vi is equal to acceleration
times time. We know that the final velocity
is equal to the initial velocity plus
acceleration times time. Let's substitute that
back into what I was writing right here. We have distance is equal to the
initial velocity plus the final velocity, so let's
substitute this expression right here. The initial velocity, plus, now
the final velocity is now the initial velocity, plus
acceleration times time, and then we divide all of that
by 2 times time. We get d is equal to-- we have
2 in the numerator, we have 2 initial velocity, 2vi's plus
at over 2, and all of that times t. Then we can simplify this. This equals d is equal to-- this
2 cancels out this 2, and then we distribute this t across
both terms-- so d is equal to vit plus-- this term
is at over 2, but then you multiply the t times here, too--
so it's at squared over 2 plus at squared over 2. We could use this formula
if we know the change in distance, or the distance-- this
actually should be the change in distance, and the
change in time-- is equal to the initial velocity times time
plus acceleration times squared divided by 2. Let me summarize all of the
equations we have, because we really now have in our arsenal
every equation that you really need to solve one dimensional
projectile problems-- things going either just left, right,
east, west, or north, south, but not both. I will do that in
the next video. Let's summarize everything
we know. We know the change in distance
divided by the change in time is equal to velocity-- average
velocity, and it would equal velocity if velocity's not
changing, but average when velocity does change-- and we
have constant acceleration, which is an important
assumption. We know that the change in
velocity divided by the change in time is equal to
acceleration. We know the average velocity is
equal to the final velocity plus the initial velocity
over 2, and this assumes acceleration is constant. If we know the initial velocity,
acceleration, and the distance, and we want to
figure out the final velocity, we could use this formula: vf
squared equals vi squared plus 2a times-- really the change in
distance, so I'm going to write the change in distance,
because that sometimes matters when we're dealing with
direction-- change in distance, but so you'll
sometimes just write this as distance. Then we just did the equation--
I think I did this in the third video, as well,
early on-- but we also learned that distance is equal to the
initial velocity times time plus at squared over 2. In that example that I did a
couple of videos ago, where we had a cliff-- actually,
I only have a minute left in this video. I will do that in the
next presentation. I'll see you soon.